Question:
Solve the following homogeneous differential equation: \( x \frac{dy}{dx} = y - x \sin^2\left(\frac{y}{x}\right) \), given the initial value condition that \( y = \frac{\pi}{6} \) when \( x = 1 \).
Solve the following homogeneous differential equation: \( x \frac{dy}{dx} = y - x \sin^2\left(\frac{y}{x}\right) \), given the initial value condition that \( y = \frac{\pi}{6} \) when \( x = 1 \).
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Whenever you see a combination like \(\sin(y/x)\), \(\cos(y/x)\), or \(e^{y/x}\), it is a clear structural indicator that the substitution \(y=vx\) is necessary. Always look out for terms that simplify immediately when \(v\) cancels from both sides!
Updated On: Jul 4, 2026
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Solution and Explanation
Concept:
The presence of the term \(\frac{y}{x}\) indicates that this is a homogeneous differential equation of degree 1. The standard method to solve such equations is:
• Substitute \( y = vx \), which implies that by using the product rule, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
• This substitution transforms the equation into a separable variable form where functions of \(v\) and functions of \(x\) can be integrated on separate sides.
Step 1: Rearrange the differential equation into standard form.
Divide both sides of the given equation by \(x\): \[ \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) \quad \cdots (1) \]
Step 2: Apply the homogeneous substitution parameters.
Let \( y = vx \). Therefore, differentiating with respect to \(x\) gives: \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute these expressions back into Equation (1): \[ v + x \frac{dv}{dx} = v - \sin^2(v) \]
Step 3: Simplify and separate variables.
Subtracting \(v\) from both sides: \[ x \frac{dv}{dx} = -\sin^2(v) \] Rearranging terms to group \(v\) terms on the left side and \(x\) terms on the right side: \[ \frac{1}{-\sin^2(v)} \, dv = \frac{1}{x} \, dx \] \[ -\csc^2(v) \, dv = \frac{1}{x} \, dx \]
Step 4: Integrate both sides.
\[ \int -\csc^2(v) \, dv = \int \frac{1}{x} \, dx \] We know that the standard integral of \( \csc^2(v) \) is \( -\cot(v) \), so \( \int -\csc^2(v) dv = \cot(v) \). The integral of \( \frac{1}{x} \) is \( \log_e|x| \): \[ \cot(v) = \log_e|x| + C \] Substituting back \( v = \frac{y}{x} \): \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + C \quad \cdots (2) \]
Step 5: Apply the boundary condition to find \(C\).
We are given that \( y = \frac{\pi}{6} \) when \( x = 1 \). Substitute these coordinates into Equation (2): \[ \cot\left(\frac{\pi/6}{1}\right) = \log_e|1| + C \] Since \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\) and \(\log_e(1) = 0\): \[ \sqrt{3} = 0 + C \quad \Rightarrow \quad C = \sqrt{3} \] Thus, the final specific solution is: \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + \sqrt{3} \]
• Substitute \( y = vx \), which implies that by using the product rule, \( \frac{dy}{dx} = v + x \frac{dv}{dx} \).
• This substitution transforms the equation into a separable variable form where functions of \(v\) and functions of \(x\) can be integrated on separate sides.
Step 1: Rearrange the differential equation into standard form.
Divide both sides of the given equation by \(x\): \[ \frac{dy}{dx} = \frac{y}{x} - \sin^2\left(\frac{y}{x}\right) \quad \cdots (1) \]
Step 2: Apply the homogeneous substitution parameters.
Let \( y = vx \). Therefore, differentiating with respect to \(x\) gives: \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \] Substitute these expressions back into Equation (1): \[ v + x \frac{dv}{dx} = v - \sin^2(v) \]
Step 3: Simplify and separate variables.
Subtracting \(v\) from both sides: \[ x \frac{dv}{dx} = -\sin^2(v) \] Rearranging terms to group \(v\) terms on the left side and \(x\) terms on the right side: \[ \frac{1}{-\sin^2(v)} \, dv = \frac{1}{x} \, dx \] \[ -\csc^2(v) \, dv = \frac{1}{x} \, dx \]
Step 4: Integrate both sides.
\[ \int -\csc^2(v) \, dv = \int \frac{1}{x} \, dx \] We know that the standard integral of \( \csc^2(v) \) is \( -\cot(v) \), so \( \int -\csc^2(v) dv = \cot(v) \). The integral of \( \frac{1}{x} \) is \( \log_e|x| \): \[ \cot(v) = \log_e|x| + C \] Substituting back \( v = \frac{y}{x} \): \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + C \quad \cdots (2) \]
Step 5: Apply the boundary condition to find \(C\).
We are given that \( y = \frac{\pi}{6} \) when \( x = 1 \). Substitute these coordinates into Equation (2): \[ \cot\left(\frac{\pi/6}{1}\right) = \log_e|1| + C \] Since \(\cot\left(\frac{\pi}{6}\right) = \sqrt{3}\) and \(\log_e(1) = 0\): \[ \sqrt{3} = 0 + C \quad \Rightarrow \quad C = \sqrt{3} \] Thus, the final specific solution is: \[ \cot\left(\frac{y}{x}\right) = \log_e|x| + \sqrt{3} \]
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